Metrically bounded set

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"Bounded set" is misused, there are other notions of bounded perhaps of sets. As such metrically bounded (set) ought to be used. This is the start of that process Alec (talk) 23:10, 18 March 2017 (UTC)

Definition

Let $(X,d)$ be a metric space and let $A\in\mathcal{P}(X)$ be an arbitrary subset of $X$ . We say $A$ is (metrically)^{[Note 1]} bounded or is a (metrically) bounded set if^[1]:

∃C∈ $\mathbb{R}_{>0}$ ∀a,b∈A[d(a,b)<C]
- In words: there exists a real $C>0$ such that for any two points in $A$ the distance between them (as measured by the metric, $d$ ) is strictly less than $C$

Note that if $A\eq\emptyset$ then it is vacuously seen to be bounded by any $C>0$ ^{[Note 2]}

Equivalent statements

A set is bounded if and only if for all points in the space there is a positive real such that the distance from that point to any point in the set is less than the positive real
- i.e. $\big(\underbrace{\forall x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C]}_{\text{Claimed condition} }\big)\iff\big(\underbrace{\exists C\in\mathbb{R}_{>0} \forall a,b\in A[d(a,b)<C]}_{\text{bounded (as above)} }\big)$
- This leads us to a useful corollary which can be found below

Implying statements

If there exists a point in the space and a positive real such that for all points in a subset of the space the distance between the selected point and the arbitrary point of the subset is less than the positive real then the subset is bounded
- $\big(\exists x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C]\big)\implies\big(\underbrace{\exists C\in\mathbb{R}_{>0} \forall a,b\in A[d(a,b)<C]}_{\text{bounded (as above)} }\big)$
- Note this is closely related to the equivalent condition above. As it allows us to pick our own point which has "bounded distance" to all points of the subset, from that we get that it is a bounded subset, and thus it would work for all points (the equivalent statement above)
  - that is to say: $\big(\exists x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C]\big)\implies\big(\exists C\in\mathbb{R}_{>0} \forall a,b\in A[d(a,b)<C]\big)\underbrace{\implies}_{(\iff)}\big(\forall x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)<C]\big)$

Notes

Jump up ↑ Alec's terminology to create a "global name" for use with pages
Jump up ↑
Proof:
- Define (choose) C:=1∈R
  - By rewriting for-all and exists within set theory we see:
    - $\forall a,b\in A[d(a,b)<C]$ is short for $\forall a\in A\forall b\in A[d(a,b)<C]$ which is short for $\forall a[a\in A\implies \forall b[b\in A\implies d(a,b)<C]]$
    - It is easy to see this is equivalent to $\forall a\forall b[(a\in A\wedge b\in A)\implies d(a,b)<C]$
    - Let $a,b$ be given
      As $A\eq\emptyset$ we cannot have $a\in A$ or $b\in A$ so the LHS of the implication is false
      By the nature of logical implication we consider it true regardless of the RHS when the LHS is false, so we're done.

References

Jump up ↑ Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha

[1] Jump up ↑ Alec's terminology to create a "global name" for use with pages

[3] Jump up ↑ Proof:
Define (choose) $C:\eq 1\in\mathbb{R}$
By rewriting for-all and exists within set theory we see:
$\forall a,b\in A[d(a,b)<C]$ is short for $\forall a\in A\forall b\in A[d(a,b)<C]$ which is short for $\forall a[a\in A\implies \forall b[b\in A\implies d(a,b)<C]]$

It is easy to see this is equivalent to $\forall a\forall b[(a\in A\wedge b\in A)\implies d(a,b)<C]$

Let $a,b$ be given
As $A\eq\emptyset$ we cannot have $a\in A$ or $b\in A$ so the LHS of the implication is false
By the nature of logical implication we consider it true regardless of the RHS when the LHS is false, so we're done.

[3] Define (choose) $C:\eq 1\in\mathbb{R}$
By rewriting for-all and exists within set theory we see:
$\forall a,b\in A[d(a,b)<C]$ is short for $\forall a\in A\forall b\in A[d(a,b)<C]$ which is short for $\forall a[a\in A\implies \forall b[b\in A\implies d(a,b)<C]]$

It is easy to see this is equivalent to $\forall a\forall b[(a\in A\wedge b\in A)\implies d(a,b)<C]$

Let $a,b$ be given
As $A\eq\emptyset$ we cannot have $a\in A$ or $b\in A$ so the LHS of the implication is false
By the nature of logical implication we consider it true regardless of the RHS when the LHS is false, so we're done.

[4] By rewriting for-all and exists within set theory we see:
$\forall a,b\in A[d(a,b)<C]$ is short for $\forall a\in A\forall b\in A[d(a,b)<C]$ which is short for $\forall a[a\in A\implies \forall b[b\in A\implies d(a,b)<C]]$

It is easy to see this is equivalent to $\forall a\forall b[(a\in A\wedge b\in A)\implies d(a,b)<C]$

Let $a,b$ be given
As $A\eq\emptyset$ we cannot have $a\in A$ or $b\in A$ so the LHS of the implication is false
By the nature of logical implication we consider it true regardless of the RHS when the LHS is false, so we're done.

[5] $\forall a,b\in A[d(a,b)<C]$ is short for $\forall a\in A\forall b\in A[d(a,b)<C]$ which is short for $\forall a[a\in A\implies \forall b[b\in A\implies d(a,b)<C]]$

[6] It is easy to see this is equivalent to $\forall a\forall b[(a\in A\wedge b\in A)\implies d(a,b)<C]$

[7] Let $a,b$ be given
As $A\eq\emptyset$ we cannot have $a\in A$ or $b\in A$ so the LHS of the implication is false
By the nature of logical implication we consider it true regardless of the RHS when the LHS is false, so we're done.

[8] As $A\eq\emptyset$ we cannot have $a\in A$ or $b\in A$ so the LHS of the implication is false
By the nature of logical implication we consider it true regardless of the RHS when the LHS is false, so we're done.

[9] By the nature of logical implication we consider it true regardless of the RHS when the LHS is false, so we're done.

[FAVIDMH-2] Jump up ↑ Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha

[Note 1]

[1]

[Note 2]

Metrically bounded set

Contents

Definition

Equivalent statements

Implying statements

See also

Notes

References

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